Question

proove the formula of escape velocity of an object of mass m thrown from the surface of earth but we have to assume zero gravitational potential energy at the surface of earth.

Answer 1

When we assume gravitational potential energy at the surface of the earth to be zero, then the gravitational potential energy ​at height h from earth ’s surface is given by,
U=mgh
$$\begin{gathered} \mathrm{Escape}\mathrm{velocity}\mathrm{is}\mathrm{the}\mathrm{velocity}\mathrm{with}\mathrm{which}\mathrm{a}\mathrm{projectile}\mathrm{has}\mathrm{to}\mathrm{be}\mathrm{projected}\mathrm{to}\mathrm{escape}\mathrm{the}\mathrm{earth}\text{'}\mathrm{s}\mathrm{gravitational}\mathrm{field}. \\ \mathrm{Suppose}\mathrm{the}\mathrm{body}\mathrm{is}\mathrm{projected}\mathrm{with}\mathrm{velocity}\text{'}\mathrm{v}\text{'},\mathrm{then} \\ \mathrm{At}\mathrm{the}\mathrm{surface},\mathrm{kinetic}\mathrm{energy},\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2 \\ \mathrm{To}\mathrm{overcome}\mathrm{the}\mathrm{earth}\text{'}\mathrm{s}\mathrm{gravitational}\mathrm{field}, \\ \mathrm{Kinetic}\mathrm{energy}>\mathrm{Potential}\mathrm{energy} \\ \mathrm{or}\frac{1}{2}{\mathrm{mv}}^2>\mathrm{mgh} \\ {\mathrm{v}}^2>2\mathrm{gh} \\ \mathrm{or}\mathrm{v}>\sqrt{2\mathrm{gh}} \\ \mathrm{Thus},\mathrm{to}\mathrm{escape}\mathrm{the}\mathrm{earth}\text{'}\mathrm{s}\mathrm{gravitational}\mathrm{field},\mathrm{the}\mathrm{minimum}\mathrm{velocity}\mathrm{must}\mathrm{be} \\ \sqrt{2\mathrm{gh}}. \end{gathered}$$